Part (i):
Let \(\varphi = \varphi (r,\xi )\)  be an arbitrary eigenfunction of (REF ) with eigenvalue \(E = E(R)\)  written in \((r,\xi )\) -spherical coordinates. Since problem (REF ) is radially symmetric, \(\varphi \)  factors as \(\varphi (r,\xi ) = u(r) Y_{\ell }(\xi )\) , where \(Y_{\ell }\)  is a spherical harmonic (eigenfunction of the Laplace–Beltrami operator on \(\mathbb {S}^{n-1}\) ) with eigenvalue \(\ell (\ell + n - 2)\)  for \(\ell \in \mathbb {N}\cup \lbrace 0\rbrace \) .
Assume that \(\ell = 0\)  so that \(\varphi (r,\xi )\)  is a constant times \(u(r)\) . In this case, \(-\Delta - \kappa /r\)  reduces to the radial part of the Laplacian plus the potential \(-\kappa /r\)  so that
\({\left\lbrace \begin{array}{ll}-u^{\prime \prime } - \frac{n-1}{r}u^{\prime } - \frac{\kappa }{r} u = E u, \quad &\text{on} ~ (0,R) \\(u^{\prime } + \gamma u)(R) = 0.\end{array}\right.}\) 
In the above problem and the ones below we impose Dirichlet boundary conditions when \(\gamma = \infty \) . Let \(m = (n-3)/2\)  and \(v(r) = r^{m+1} u(r)\)  so that
\({\left\lbrace \begin{array}{ll}-v^{\prime \prime } + \frac{m(m + 1)}{r^2}v - \frac{\kappa }{r} v = E v, \quad &\text{on} ~ (0,R) \\(v^{\prime } +(\gamma - \frac{m+1}{R})v)(R) = 0.\end{array}\right.}\) 
Since \(E \ne 0\)  define \( z = \kappa (2\sqrt{|E|})^{-1} > 0\)  and rescale the radial coordinate by defining \(\rho = \kappa r/ z\)  and let \(w(\rho ) = v( z \rho /\kappa )\) . It follows that \(w\)  satisfies
\({\left\lbrace \begin{array}{ll}-w^{\prime \prime } + \frac{m(m + 1)}{\rho ^2}w - (\frac{ z}{\rho } - \frac{1}{4})w = 0, \quad &\text{on} ~ (0,\rho _0) \\(w^{\prime } + (\frac{ z \gamma }{\kappa } - \frac{m+1}{\rho _0})w)(\rho _0) = 0,\end{array}\right.}\) 
where \(\rho _0 = \kappa R/ z\) .
Guessing a solution of the form \(w(\rho ) = \rho ^{m + 1} e^{-\rho /2} F(\rho )\)  we find that \(F(\rho ) = F(a_0,b_0;\rho )\)  is a solution of (REF ) on the interval \((0,\rho _0)\)  with \(a_0(z) = m +1 - z\) , \(b = b_0 = 2m + 2\) , and boundary condition \(2F^{\prime }(a_0,b_0;\rho _0) = (1 - z \gamma /\kappa )F(a_0,b_0;\rho _0)\) . When \(\gamma = \infty \)  the boundary condition \(F(a_0,b_0;\rho _0) = 0\)  is immediate from the form of the ansatz. Thus, the original radial eigenfunctions \(\varphi \)  are multiples of \(F(a_0,b_0;2\sqrt{|E|}r)e^{-\sqrt{|E|}r}\) .
The general solution of
(REF ) is a linear combination of the confluent hypergeometric functions of the first and second kind, \(M(a_0,b_0;\rho )\) and \(U(a_0,b_0;\rho )\) , respectively. It follows from formulas 13.4.21, 13.5.5, and 13.5.7 in [1]} that
\(U^{\prime }(a_0,b_0;\rho ) = -a_0\frac{\Gamma (b_0)}{\Gamma (a_0+1)}\rho ^{-b_0} + o(\rho ^{-b_0}),\quad \text{as } \rho \rightarrow 0,\) 
so that \(U^{\prime }(a_0,b_0;\rho ) \notin L^2(B(R))\)  since \(b_0 = n-1\) . Thus, \(F(\rho )\)  must have no \(U(a_0,b_0;\rho )\)  component to guarantee that \(\varphi \in H^1(B(R))\) . This shows that \(F(\rho ) = M(a_0,b_0;\rho )\) .
Finally, using the boundary condition we computed for \(F\)  and the formula \(M^{\prime }(a,b;\rho ) = \frac{a}{b}M(a+1,b+1;\rho )\)  [1]} we deduce that \(z\)  must satisfy the transcendental equation (REF ). Conversely, each positive solution \(z\)  to the transcendental equation defines an eigenvalue \(E = \operatorname{sgn}(E)\kappa ^2/4 z^2\)  corresponding to a radial eigenfunction of (REF ). Thus, the eigenvalues \(E\)  corresponding to radial eigenfunctions of (REF ) are in one-to-one correspondence with solutions \(z\)  of (REF ), and part (i) of the lemma on \(B(R)\)  follows from this correspondence.
The same calculation shows that on \(\mathbb {R}^n\)  the eigenfunctions are multiples of (REF ) with eigenvalue \(-\kappa ^2/4z^2\)  and that \(w(\rho ) = \rho ^{m+1} e^{-\rho /2}M(a_0,b_0;\rho )\) . Since we are assuming that the original eigenfunction \(\varphi \)  tends to zero as \(r \rightarrow \infty \) , the function \(w\)  grows at most polynomially. If \(a\)  and \(b\)  are fixed and \(a \notin \lbrace 0,-1,-2,\dots \rbrace \)  then \(M(a,b;\rho ) \sim \rho ^{a-b}e^\rho \)  as \(\rho \rightarrow \infty \)  (see [1]}). If \(a \in \lbrace 0,-1,-2,\dots \rbrace \)  then \(M(a,b;\rho )\)  is a polynomial of degree \(-a\)  (see (REF )). Thus, we must have \(a_0 \in \lbrace 0,-1,-2,\dots \rbrace \)  to ensure \(\varphi \)  tends to zero as \(r \rightarrow \infty \) . It follows from the definition of \(a_0 = m+1 - z\)  that \(z = \kappa (2 \sqrt{|E|})^{-1}\)  and \(E = -\kappa ^2/4(j + (n - 3)/2)^2\)  for \(j \in \mathbb {N}\) .
Part (ii): We split the the proof into two cases depending on the size of \(|\gamma |\) . Recall that \(\kappa > 0\)  and \(m = (n-3)/2\) .
Case 1: \(|\gamma | < \kappa /(2(m+1))\) : We show that the smallest positive \(z = z(\rho _0)\)  that solves (REF ) exists and tends to \(m + 1\)  from below exponentially fast, as \(\rho _0 \rightarrow \infty \)  (equivalently as \(R \rightarrow \infty \)  in our original variables). We do this by inverting this problem to show that there is a \(\rho _0 = \rho _0(z)\)  that solves (REF ) exists and tends to infinity, as \(z \nearrow m + 1\) . First we fix \(c > 1\)  and \(0 < z < m+1\)  close to \(m + 1\)  and apply the intermediate value theorem to the function
\(\mathcal {M}(\rho ) = 2\frac{a_0}{b_0} M(a_0 + 1,b_0 + 1; \rho ) - (1 - 2\gamma z/\kappa )M(a_0,b_0; \rho ).\) 
Since \(\mathcal {M}(\rho ) = 0\)  if and only if \(\rho \)  satisfies the transcendental equation, we show that for \(c\)  and \(z\)  as above there exists a \(\rho _0(z)\)  such that \(\mathcal {M}(\rho _0(z)) = 0\)  and
\(\rho _-(z) \le \rho _0(z) \le \rho _+(z),\) 
where
\(\rho _-(z) = \log (1 - 2\gamma z/\kappa ) + \log \bigg (\frac{m + 1}{m + 1 - z} \bigg ) \quad \text{and} \quad \rho _+(z) = c \log \bigg (\frac{m + 1}{m + 1 - z} \bigg ).\) 
(Note we need \(z\)  large enough so that \(0 < \rho _-(z) < \rho _+(z)\) ). The desired convergence for \(E(R)\)  as \(R \rightarrow \infty \)  follows from solving these inequalities for \(z\)  and using the definition of \(z\)  and \(\rho _0\) .
We prove (REF ) by showing that \(\mathcal {M}(\rho )< 0\)  for \(\rho < \rho _-(z)\)  and that \(\mathcal {M}(\rho ) > 0\)  for \(\rho > \rho _+(z)\)  in order to apply the intermediate value theorem. To prove these estimates on \(\mathcal {M}\)  we use the power series representation
\(M(a,b;\rho ) = \sum _{k = 0}^{\infty } \frac{a^{(k)}}{b^{(k)}} \frac{\rho ^k}{k!}, \quad \text{for } a \in \mathbb {R},~ b \notin \lbrace 0,-1,-2,\dots \rbrace ,\) 
where \(a^{(k)}\)  is the rising factorial. That is, \(a^{(0)} = 1\)  and \(a^{(k)} = a(a+1) \cdots (a + k - 1)\)  for \(k \ge 1\) . Observe that the intermediate value theorem applies because \(\rho \mapsto M(a,b;\rho )\)  is analytic.
Since \(z < m+1\)  we know \(M(a_0,b_0;~\cdot ~)\)  is positive due to (REF ). Thus, as soon as we prove that \(\rho _0(z)\)  exists for some \(0 < z < m+1\)  we know that it corresponds to the first eigenvalue \(E\)  of (REF ) since the ground state is the unique positive eigenfunction.
First we prove \(\mathcal {M}(\rho ) < 0\)  when \(0 < \rho < \rho _-(z)\) . Using that \(a_0 + 1 = m + 2 - z < 2 m + 3 = b_0 + 1\)  we have
\(M(a_0 + 1, b_0 + 1;\rho ) < \sum _{k = 0}^{\infty } \frac{\rho ^k}{k!} = e^{\rho }.\) 
From this estimate and that \(\rho < \rho _-(z)\)  we have
\(2\frac{a_0}{b_0} M(a_0 + 1,b_0 + 1; \rho ) < \frac{m + 1 - z}{m + 1} e^{ \rho } < (1 - 2 \gamma z/ \kappa ) < (1 - 2 \gamma z/ \kappa )M(a_0, b_0; \rho ),\) 
since \(M(a_0,b_0;0) = 1\)  and \(\rho \mapsto M(a_0,b_0;\rho )\)  is increasing. Thus, \(\mathcal {M}(\rho ) < 0\)  when \(\rho < \rho _-(z)\) .
Now we prove \(\mathcal {M}(\rho ) > 0\)  when \(\rho > \rho _+(z)\)  and \(z < m+1\)  is sufficiently close to \(m+1\) . First observe that for each \(\delta \in (0,1)\)  there is a \(K\)  independent of \(z\)  such that
\(2\frac{k + m + 1 - z}{k + 2m + 2} \ge 1 + \delta , \quad \text{for each } k \ge K.\) 
The \(k^{\text{th}}\)  coefficient of the power series of the first term of \(\mathcal {M}\)  satisfies the following identity by considering the first and last factors in the rising factorials
\(2 \frac{a_0}{b_0} \frac{(a_0 + 1)^{(k)}}{(b_0 + 1)^{(k)}} = 2\frac{m + 1 - z}{2m + 2}\frac{(m + 2 - z)^{(k)}}{(2 m + 3)^{(k)}} = 2\frac{k + m + 1 - z}{k + 2m + 2} \frac{(m + 1 - z)^{(k)}}{(2 m + 2)^{(k)}},\) 
for each \(k \ge 0\) . Applying the lower bound (REF ) we have
\(2 \frac{a_0}{b_0} \frac{(a_0 + 1)^{(k)}}{(b_0 + 1)^{(k)}} \ge (1 + \delta ) \frac{a_0^{(k)}}{b_0^{(k)}}, \quad \text{for each } k \ge K.\) 
Dropping the first \(K\)  terms in the power series of \(M(a_0 + 1,b_0 + 1;\rho )\) , applying (REF ), and adding and subtracting \(1 + \delta \)  times the first \(K\)  terms in the series for \(M(a_0,b_0;\rho )\)  shows
\(2\frac{a_0}{b_0} M(a_0 + 1,b_0 + 1; \rho ) \ge (1 + \delta ) \sum _{k = K}^{\infty } \frac{a_0^{(k)}}{b_0^{(k)}} \frac{\rho ^k}{k!} = (1 + \delta ) M(a_0,b_0;\rho ) - (1 + \delta ) \sum _{k = 0}^{K - 1} \frac{a_0^{(k)}}{b_0^{(k)}} \frac{\rho ^k}{k!}. \nonumber \) 

Subtracting and adding \((2\gamma z/\kappa )M(a_0,b_0;\rho )\)  to the right side shows
\(2\frac{a_0}{b_0} M(a_0 + 1,b_0 + 1; \rho ) \ge (1 - 2\gamma z/\kappa ) M(a_0,b_0;\rho ) + \mathcal {E}(\rho ),\) 
where
\(\mathcal {E}(\rho ) = (\delta + 2\gamma z/\kappa ) M(a_0,b_0; \rho ) - (1 + \delta ) \sum _{k = 0}^{K - 1} \frac{(m + 1 - z)^{(k)}}{(2 m + 2)^{(k)}} \frac{\rho ^k}{k!}.\) 
To conclude that \(\mathcal {M}(\rho ) > 0\)  from (REF ) we will show that \(\mathcal {E}(\rho ) > 0\)  for \(\rho \ge \rho _+(z)\) , as \( z \nearrow m + 1\) , by reducing to showing that \(\mathcal {E}(\rho _+(z)) > 0\) . First, notice that for \(\delta \)  sufficiently close to 1 the factor \(\delta + 2\gamma z/\kappa \)  is positive since \(\gamma > -\kappa /(2(m+1))\) . Consequently, \(\rho \mapsto \mathcal {E}(\rho )\)  is an increasing function for fixed \(z\)  and large \(\rho \) . In particular, the first term of \(\mathcal {E}\)  is bounded from below by \(C\rho ^{K}\)  for large \(\rho \) , where \(C > 0\)  is independent of \(\rho \) , and the second term is only a polynomial of degree \(K-1\) . Thus, it suffices to prove \(\mathcal {E}(\rho _+(z)) > 0\) , as \( z \nearrow m+1\) . Since the second term of \(\mathcal {E}(\rho _+(z))\)  behaves like \(\log (1/(m + 1 - z))^{K - 1}\) , as \( z \nearrow m + 1\) , it is enough to show that \(M(a_0,b_0;\rho _+(z))\)  grows like a negative power of \(1 + m - z\) , as \( z \nearrow m + 1\) .
We estimate the factor \(a_0^{(k)}/b_0^{(k)}\)  from below using that \(m + 1 - z > 0\)  and converting the rising factorials to ordinary factorials so that for \(k \ge 1\)
\(\frac{a_0^{(k)}}{b_0^{(k)}} = \frac{(m + 1 - z)^{(k)}}{(2 m + 2)^{(k)}} \ge (m + 1 - z) \frac{(k-1)!}{(2 m + 2)^{(k)}} = (m + 1 - z) \frac{(k-1)!(2m + 1)!}{(2 m + 2 + (k-1))!}.\) 
In what follows \(C\)  is a positive constant that is independent of both \(k\)  and \(z\)  that may change from line to line. Noting that \((2m + 1)!\)  only depends on the dimension \(n\)  and making cancellations leads to the lower bound
\(\frac{a_0^{(k)}}{b_0^{(k)}} \ge (m + 1 - z) \frac{C}{(k + 2 m + 1)^{2 m + 2}}, \quad \text{for } k \ge 0.\) 
Using the above lower bound, factoring \(c = c^{1/2}c^{1/2} > 1\) , and using the Taylor series for \(e^\rho \)  we have
\(&M(a_0,b_0;\rho _+(z)) \nonumber \\\ge &C(m + 1 - z) \sum _{k = 0}^{\infty } \bigg ( \frac{c^{k/2}}{(k + 2m + 1)^{2m+2}} \bigg ) \frac{1}{k!} [c^{1/2}\log ((m + 1)/(m + 1 - z))]^k \nonumber \\\ge &C(m + 1 - z) \sum _{k = 0}^{\infty } \frac{1}{k!} [c^{1/2}\log ((m + 1)/(m + 1 - z))]^k \nonumber \\\ge &\frac{C}{(m + 1 - z)^{c^{1/2}-1}}. \nonumber \) 

Since \(c > 1\)  we know \(M(a_0,b_0;\rho _+(z))\)  blows up algebraically so that \(\mathcal {E}(\rho _+(z))\)  is positive as \( z \nearrow m + 1\) . Thus, \(\mathcal {M}(\rho ) > 0\)  for all \(\rho \ge \rho _+(z)\)  and \(z < m+1\)  sufficiently large.
Since \(\rho \mapsto \mathcal {M}(a,b;\rho )\)  is continuous there exists a solution \(\rho _0 = \rho _0( z)\)  to (REF ) that satisfies the inequalities (REF ) by the intermediate value theorem. Exponentiating these inequalities, using that \(1 - 2\gamma z/\kappa \)  is uniformly positive in \(z\) , solving for \(z\) , and using the definition \(\rho _0 = \kappa R/ z\)  we find that for \(R\)  sufficiently large there are constants \(C_1,C_2 > 0\)  such that
\(m + 1 - C_1 e^{-\kappa R/c z} \le z \le m + 1 - C_2 e^{- \kappa R/ z}.\) 
Using \(z < m + 1 = (n-1)/2\)  and that \(c > 1\)  is arbitrary we that
\(\frac{n-1}{2} - C_1 \exp \bigg \lbrace -(1 - \epsilon )\frac{2\kappa }{n-1} R\bigg \rbrace  \le z \le \frac{n-1}{2} + C_2 \exp \bigg \lbrace -(1 - \epsilon )\frac{2\kappa }{n-1} R\bigg \rbrace ,\) 
for all \(R\)  sufficiently large and for each \(\epsilon > 0\) . This shows the first estimate in part (ii) of the lemma for \(|\gamma | < \kappa /(2(m+1))\) .
From the definition \(z = \kappa (2|E|^{1/2})^{-1}\)  and that \(E < 0\)  we find that
\(E(R) = -\kappa ^2/(n-1)^2 + O\bigg (\exp \bigg \lbrace -(1 - \epsilon )\frac{2 \kappa }{n-1} R\bigg \rbrace \bigg ), \quad \text{as } R \rightarrow \infty , \quad \text{for each } \epsilon > 0,\) 
by linearizing the reciprocal function. This completes the proof of part (ii) of the lemma for \(|\gamma | < \kappa /(2(m+1))\) .
Case 2: \(\gamma \in [\kappa /(2(m+1)),\infty ]\) : In the remainder of the proof let \(E(R,\gamma )\)  denote the first eigenvalue of the Schrödinger problem (REF ). To prove the asymptotic for \(E(R,\gamma )\)  when \(\gamma \in [\kappa /(2(m+1)),\infty ]\)  notice that \(\gamma \mapsto E(R,\gamma )\)  is increasing for \(\gamma \in (-\infty ,\infty ]\)  by the min-max principle. In particular, \(E(R,0) \le E(R,\gamma ) \le E(R,\infty )\)  for each \(\gamma > 0\) . Thus, the right side of (REF ) is a lower bound for \(E(R,\gamma )\)  and it suffices to prove it is also an upper bound for the first Dirichlet eigenvalue \(E(R,\infty )\) .
By part (i) of the lemma, the ground state of the \(-\Delta - \kappa /r\)  on \(\mathbb {R}^n\)  is \(\exp \lbrace -\sqrt{|E(\infty )|}r\rbrace \)  with eigenvalue \(E(\infty ) = -\kappa ^2/(n-1)^2\) . Let \(\chi \)  be a smooth radial cutoff function that is 1 for \(r < 1 - \epsilon \) , decreases to zero for \(1 - \epsilon \le r \le 1\) , and is zero for \(r > 1\) . The function \(f(r) = \chi (r/R) \exp \lbrace -\sqrt{|E(\infty )|}r\rbrace \)  restricted to the ball \(B(R)\)  is a valid trial function for the first eigenvalue \(E(R,\infty )\)  of (REF ) so that
\(E(R,\infty ) &\le \frac{\int _{B(R)} f(-\Delta - \kappa /r)f \, dV}{\int _{B(R)} f^2 \, dV}\nonumber \\&= \frac{E(\infty )\int _{B((1 - \epsilon )R)} \exp \lbrace -2\sqrt{|E(\infty )|}r\rbrace  \, dV + \int _{B(R) \setminus B((1 - \epsilon )R)} f(-\Delta - \kappa /r)f \, dV}{\int _{B((1 - \epsilon )R)} \exp \lbrace -2\sqrt{|E(\infty )|}r\rbrace  \, dV + \int _{B(R) \setminus B((1 - \epsilon )R)} f^2 \, dV}\nonumber \\&= E(\infty ) + O\bigg (\exp \lbrace -(1 - \epsilon )\frac{2 \kappa }{n-1} R\rbrace \bigg ), \quad \text{as } R \rightarrow \infty , \quad \text{for each } \epsilon > 0, \nonumber \) 

The last equality follows from using the decay of \(\exp \lbrace -\sqrt{|E(\infty )|}r\rbrace \)  and that \(\chi (r/R)\)  has bounded \(r\) -derivatives. These facts show that the second terms in the numerator and denominator are the same size as the error term in (REF ) since \(E(\infty ) = -\kappa ^2/(n-1)^2\) .
This completes the proof of the asymptotic for \(E(R,\gamma )\)  for all \(\gamma \in (-\kappa /(2(m+1)),\infty ]\) . Note that the asymptotic for \(z\)  in part (ii) of the lemma follows from the formula \(z = \kappa (2 \sqrt{|E(R,\gamma )|})^{-1}\) .
